We have \(A^\circ \subseteq A\) and \(B^\circ \subseteq B\) and therefore \(A^\circ \cap B^\circ \subseteq A \cap B\). As \(A^\circ \cap B^\circ\) is open we then have \(A^\circ \cap B^\circ \subseteq (A \cap B)^\circ\) because \(A^\circ \cap B^\circ\) is open and \((A \cap B)^\circ\) is the largest open subset of \(A \cap B\).

For the first one, let’s take for \(E\) the plane \(\mathbb R^2\) endowed with usual topology. For \(A\), we take the unit close disk and for \(B\) the plane minus the open unit disk. \(A^\circ\) is the unit open disk and \(B^\circ\) the plane minus the unit closed disk. Therefore \(A^\circ \cup B^\circ = \mathbb R^2 \setminus C\) is equal to the plane minus the unit circle \(C\). While we have \[A \cup B = (A \cup B)^\circ = \mathbb R^2.\]

Let’s introduce and describe some properties of the line with two origins.

Let \(X\) be the union of the set \(\mathbb R \setminus \{0\}\) and the two-point set \(\{p,q\}\). The line with two origins is the set \(X\) topologized by taking as base the collection \(\mathcal B\) of all open intervals in \(\mathbb R\) that do not contain \(0\), along with all sets of the form \((-a,0) \cup \{p\} \cup (0,a)\) and all sets of the form \((-a,0) \cup \{q\} \cup (0,a)\), for \(a > 0\).

\(\mathcal B\) is a base for a topology \(\mathcal T\) of \(X\)

Indeed, one can verify that the elements of \(\mathcal B\) cover \(X\) as \[
X = \left( \bigcup_{a > 0} (-a,0) \cup \{p\} \cup (0,a) \right) \cup \left( \bigcup_{a > 0} (-a,0) \cup \{q\} \cup (0,a) \right)\] and that the intersection of two elements of \(\mathcal B\) is the union of elements of \(\mathcal B\) (verification left to the reader).

Each of the spaces \(X \setminus \{p\}\) and \(X \setminus \{q\}\) is homeomorphic to \(\mathbb R\)

Consider a normed vector space \((X, \Vert \cdot \Vert)\). If \(X\) is finite-dimensional, then a subset \(Y \subset X\) is compact if and only if it is closed and bounded. In particular a closed ball \(B_r[a] = \{x \in X \, ; \, \Vert x – a \Vert \le r\}\) is always compact if \(X\) is finite-dimensional.

The first one is based on open covers. For \(n \ge 1\), we denote by \(f_n\) the piecewise linear map defined by \[
\begin{cases}
f_n(0)=f_n(\frac{1}{2^n}-\frac{1}{2^{n+2}})=0 \\
f_n(\frac{1}{2^n})=1 \\
f_n(\frac{1}{2^n}+\frac{1}{2^{n+2}})=f_n(1)=0
\end{cases}\] All the \(f_n\) belong to \(B_1[0]\). Moreover for \(1 \le n < m\) we have \(\frac{1}{2^n}+\frac{1}{2^{n+2}} < \frac{1}{2^m}-\frac{1}{2^{m+2}}\). Hence the supports of the \(f_n\) are disjoint and \(\Vert f_n – f_m \Vert = 1\).

Now consider the open cover \(\mathcal U=\{B_{\frac{1}{2}}(x) \, ; \, x \in B_1[0]\}\). For \(x \in B_1[0]\}\) and \(u,v \in B_{\frac{1}{2}}(x)\), \(\Vert u -v \Vert < 1\). Therefore, each \(B_{\frac{1}{2}}(x)\) contains at most one \(f_n\) and a finite subcover of \(\mathcal U\) will contain only a finite number of \(f_n\) proving that \(A\) is not compact.

Second proof based on convergent subsequence. As \(A\) is a metric space, it is enough to prove that \(A\) is not sequentially compact. Consider the sequence of functions \(g_n : x \mapsto x^n\). The sequence is bounded as for all \(n \in \mathbb N\), \(\Vert g_n \Vert = 1\). If \((g_n)\) would have a convergent subsequence, the subsequence would converge pointwise to the function equal to \(0\) on \([0,1)\) and to \(1\) at \(1\). As this function is not continuous, \((g_n)\) cannot have a subsequence converging to a map \(g \in A\).

Riesz’s theorem

The non-compactness of \(A=C([0,1],\mathbb R)\) is not so strange. Based on Riesz’s lemma one can show that the unit ball of an infinite-dimensional normed space \(X\) is never compact. This is sometimes known as the Riesz’s theorem.

The non-compactness of \(A=C([0,1],\mathbb R)\) is just standard for infinite-dimensional normed vector spaces!

An open ball \(B_r(p)\) whose closure is not equal to the closed ball \(B_r[p]\)

Here we take for \(M\) a subspace of \(\mathbb R^2\) which is the union of the origin \(\{0\}\) with the unit circle \(S^1\). For the distance, we use the Euclidean norm.
The open unit ball centered at the origin \(B_1(0)\) is reduced to the origin: \(B_1(0) = \{0\}\). Its closure \(\overline{B_1(0)}\) is itself. However the closed ball \(B_1[0]\) is the all space \(\{0\} \cup S^1\).

Again one can prove that for a normed vector space this cannot happen. The closure of an open ball is the closed ball for a normed vector space.

In this article we look at what happens to Banach-Steinhaus theorem when the completness hypothesis is not fulfilled. One form of Banach-Steinhaus theorem is the following one.
Banach-Steinhaus Theorem Let \(T_n : E \to F\) be a sequence of continuous linear maps from a Banach space \(E\) to a normed space \(F\). If for all \(x \in E\) the sequence \(T_n x\) is convergent to \(Tx\), then \(T\) is a continuous linear map.

A sequence of continuous linear maps converging to an unbounded linear map

A connected space is a topological space that cannot be represented as the union of two or more disjoint nonempty open subsets. We look here at unions and intersections of connected spaces.

Union of connected spaces

The union of two connected spaces \(A\) and \(B\) might not be connected “as shown” by two disconnected open disks on the plane.The union of two connected spaces might not be connected.
However if the intersection \(A \cap B\) is not empty then \(A \cup B\) is connected.

Intersection of connected spaces

The intersection of two connected spaces \(A\) and \(B\) might also not be connected. An example is provided in the plane \(\mathbb R^2\) by taking for \(A\) the circle centered at the origin with radius equal to \(1\) and for \(B\) the segment \(\{(x,0) \ : \ x \in [-1,1]\}\). The intersection \(A \cap B = \{(-1,0),(1,0)\}\) is the union of two points which is not connected.